= Solution
Model the two samples as independent multinomial counts with fixed row totals $N_A=N_B=500$ and three category probabilities. The null hypothesis is $p_{Aj}=p_{Bj}=p_j$ for every category $j$. Up to combinatorial factors the <likelihood> is $\prod_{ij}p_{ij}^{O_{ij}}$. Without the null restriction its maximizers are $O_{ij}/N_i$; under the null, maximizing $\sum_j(O_{Aj}+O_{Bj})\log p_j$ subject to $\sum_jp_j=1$ gives
$$
\widehat p_j=\frac{O_{Aj}+O_{Bj}}{1000},\qquad
E_{ij}=N_i\widehat p_j.
$$
The observed column totals are $243,281,476$, giving fitted expected counts \b[$(121.5,140.5,238)$ in each row].
To derive the null distribution, put $D=O_A-O_B$. Under the null its mean is zero and covariance is $1000(\operatorname{diag}p-pp^T)$. The multivariate <central limit theorem> gives, asymptotically, the standardized contrast $Z_j=D_j/\sqrt{1000p_j}$ a centered <normal distribution> with covariance
$$
I-uu^T,\qquad u=(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})^T,\quad u^Tu=1.
$$
This is the <orthogonal projection> onto a two-dimensional plane. In an orthonormal <basis> of that plane, the two nonzero coordinates are independent standard normal variables, so $\sum_jZ_j^2$ tends to $\chi_2^2$. The pooled estimates are consistent; substituting them for $p_j$ leaves this limit unchanged by <Slutsky's theorem>. Since $O_{Aj}-E_{Aj}=D_j/2$ and $O_{Bj}-E_{Bj}=-D_j/2$, the resulting statistic is precisely the <Pearson chi-squared test of homogeneity> statistic
$$
X^2=\sum_{i,j}\frac{(O_{ij}-E_{ij})^2}{E_{ij}}
=\sum_j\frac{D_j^2}{1000\widehat p_j},\qquad X^2\xrightarrow{d}\chi_2^2.
$$
This supplies the two degrees of freedom rather than treating all six cells as independent. Every fitted expected count is large, supporting the approximation.
For these data,
$$
\boxed{X^2=2\left(\frac{18.5^2}{121.5}+\frac{4.5^2}{140.5}
+\frac{14^2}{238}\right)=7.56906.}
$$
It exceeds the two-degree-of-freedom 95th percentile $5.99$, but not the 99th percentile $9.21$. Hence \b[reject equal score distributions at the 5% level, but not at the 1% level]. The asymptotic p-value is $\Pr(\chi_2^2\ge7.56906)=e^{-7.56906/2}\simeq0.02272$.
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